Asked by Shelly
Question: A(n) 3.17 m long pendulum is released from
rest when the support string is at an angle of 38.6 degrees with the vertical.
The acceleration of gravity is 9.8 m/s2 .
What is the speed of the bob at the bottom
of the swing?
Answer in units of m/s
What I did:
y=3.17cos38.6 = 1.968297399
h=3.17m-1.968297399m =1.20170261m
v^2 =2(9.8)(1.20170261) =23.55337116
sqrt(23.55337116) =4.85318155
what did I do wrong and how do I fix it? thanks
rest when the support string is at an angle of 38.6 degrees with the vertical.
The acceleration of gravity is 9.8 m/s2 .
What is the speed of the bob at the bottom
of the swing?
Answer in units of m/s
What I did:
y=3.17cos38.6 = 1.968297399
h=3.17m-1.968297399m =1.20170261m
v^2 =2(9.8)(1.20170261) =23.55337116
sqrt(23.55337116) =4.85318155
what did I do wrong and how do I fix it? thanks
Answers
Answered by
Elena
y=3.17cos38.6 =3.17 •0.78=2.48
h= 3.17 - 2.48=0.69
v=3.68 m/s
h= 3.17 - 2.48=0.69
v=3.68 m/s
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